The Electric Field Formula gives the force experienced per unit positive charge at any point in space, expressed as \ ( E = \frac{kQ}{r^2} \), where E is the electric field intensity in N/C. This fundamental concept is covered in Class 12 Physics, Chapter 1 (Electric Charges and Fields) of the NCERT textbook. It is equally vital for JEE Main, JEE Advanced, and NEET examinations, where electrostatics questions carry significant weightage. This article covers the complete formula, derivation, a full formula sheet, three solved examples at progressive difficulty levels, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Electric Field Formulas at a Glance
Quick reference for the most important Electric Field Formula expressions used in CBSE Class 12 and competitive exams.
- Electric field due to a point charge: \( E = \frac{kQ}{r^2} = \frac{Q}{4\pi\epsilon_0 r^2} \)
- Electric force on a charge: \( F = qE \)
- Electric field between parallel plates: \( E = \frac{\sigma}{\epsilon_0} \)
- Electric field due to a dipole (axial): \( E = \frac{2kp}{r^3} \)
- Electric field due to a dipole (equatorial): \( E = \frac{kp}{r^3} \)
- Gauss's Law: \( \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} \)
- Electric field of a uniformly charged sphere (outside): \( E = \frac{Q}{4\pi\epsilon_0 r^2} \)
What is Electric Field?
The Electric Field Formula describes a physical quantity that represents the electric force per unit positive test charge at a given point in space. An electric field exists around every charged object. It is a vector quantity, meaning it has both magnitude and direction.
According to NCERT Class 12 Physics, Chapter 1 — Electric Charges and Fields, the electric field at a point is defined as the force that a unit positive charge would experience if placed at that point. The direction of the electric field is the direction of force on a positive charge.
The concept was introduced by Michael Faraday to explain how charges interact without direct contact. The field permeates the space around the source charge and exerts a force on any other charge placed within it. Understanding this concept is essential for mastering topics like capacitance, Gauss's Law, and electromagnetic induction in higher classes.
The SI unit of electric field is Newton per Coulomb (N/C), which is equivalent to Volt per metre (V/m). The electric field is always directed away from a positive source charge and towards a negative source charge.
Electric Field Formula — Expression and Variables
The standard Electric Field Formula for a point charge is:
\[ E = \frac{kQ}{r^2} = \frac{Q}{4\pi\epsilon_0 r^2} \]
where k is Coulomb's constant, equal to \( 9 \times 10^9 \, \text{N m}^2 \text{C}^{-2} \), and \( \epsilon_0 \) is the permittivity of free space, equal to \( 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \).
| Symbol | Quantity | SI Unit |
|---|---|---|
| E | Electric Field Intensity | N/C or V/m |
| k | Coulomb's Constant (\( 9 \times 10^9 \)) | N m² C⁻² |
| Q | Source Charge | Coulomb (C) |
| r | Distance from source charge to the point | Metre (m) |
| \( \epsilon_0 \) | Permittivity of Free Space | C² N⁻¹ m⁻² |
| F | Electrostatic Force on test charge | Newton (N) |
| q | Test Charge | Coulomb (C) |
Derivation of Electric Field Formula
The Electric Field Formula is derived from Coulomb's Law. Consider a source charge Q and a small positive test charge q placed at a distance r from Q.
Step 1: By Coulomb's Law, the force between the two charges is:
\[ F = \frac{kQq}{r^2} \]
Step 2: The electric field E at that point is defined as the force per unit positive test charge:
\[ E = \frac{F}{q} \]
Step 3: Substituting the expression for F:
\[ E = \frac{kQq}{r^2 \cdot q} = \frac{kQ}{r^2} \]
The test charge q cancels out, showing that the electric field depends only on the source charge Q and the distance r. This confirms that E is a property of the source charge and the location in space, independent of the test charge used to measure it.
Complete Electrostatics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Electric Field (Point Charge) | \( E = \frac{kQ}{r^2} \) | k = Coulomb's constant, Q = charge, r = distance | N/C | Class 12, Ch 1 |
| Coulomb's Law | \( F = \frac{kQ_1Q_2}{r^2} \) | Q₁, Q₂ = charges, r = separation | Newton (N) | Class 12, Ch 1 |
| Electric Force on Charge | \( F = qE \) | q = charge in field E | Newton (N) | Class 12, Ch 1 |
| Electric Field (Dipole, Axial) | \( E_{axial} = \frac{2kp}{r^3} \) | p = dipole moment, r = distance from centre | N/C | Class 12, Ch 1 |
| Electric Field (Dipole, Equatorial) | \( E_{eq} = \frac{kp}{r^3} \) | p = dipole moment, r = distance from centre | N/C | Class 12, Ch 1 |
| Electric Flux | \( \Phi = E \cdot A \cdot \cos\theta \) | A = area, θ = angle with normal | N m²/C | Class 12, Ch 1 |
| Gauss's Law | \( \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} \) | Q_enc = enclosed charge | N m²/C | Class 12, Ch 1 |
| Electric Field (Infinite Plane Sheet) | \( E = \frac{\sigma}{2\epsilon_0} \) | σ = surface charge density | N/C | Class 12, Ch 1 |
| Electric Field (Between Parallel Plates) | \( E = \frac{\sigma}{\epsilon_0} \) | σ = surface charge density | N/C | Class 12, Ch 2 |
| Electric Potential Energy | \( U = \frac{kQ_1Q_2}{r} \) | Q₁, Q₂ = charges, r = separation | Joule (J) | Class 12, Ch 2 |
| Electric Potential | \( V = \frac{kQ}{r} \) | Q = source charge, r = distance | Volt (V) | Class 12, Ch 2 |
| Relation: E and V | \( E = -\frac{dV}{dr} \) | V = potential, r = distance | V/m | Class 12, Ch 2 |
Solved Examples on Electric Field Formula
Example 1 (Class 9-10 Level — Direct Application)
Problem: Calculate the electric field intensity at a point 0.3 m away from a point charge of 5 μC in free space.
Given: Q = 5 μC = 5 × 10⁻⁶ C, r = 0.3 m, k = 9 × 10⁹ N m² C⁻²
Step 1: Write the Electric Field Formula: \( E = \frac{kQ}{r^2} \)
Step 2: Substitute the values: \( E = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{(0.3)^2} \)
Step 3: Calculate the denominator: \( (0.3)^2 = 0.09 \, \text{m}^2 \)
Step 4: Calculate the numerator: \( 9 \times 10^9 \times 5 \times 10^{-6} = 45 \times 10^3 = 4.5 \times 10^4 \)
Step 5: Divide: \( E = \frac{4.5 \times 10^4}{0.09} = 5 \times 10^5 \, \text{N/C} \)
Answer
The electric field intensity = 5 × 10⁵ N/C, directed away from the positive charge.
Example 2 (Class 11-12 Level — Multi-Step)
Problem: Two point charges, Q₁ = +4 μC and Q₂ = −4 μC, are placed 20 cm apart. Find the electric field at the midpoint of the line joining the two charges.
Given: Q₁ = +4 μC = 4 × 10⁻⁶ C, Q₂ = −4 μC = −4 × 10⁻⁶ C, separation = 20 cm = 0.20 m, midpoint is at r = 0.10 m from each charge.
Step 1: Find E₁ at midpoint due to Q₁:
\( E_1 = \frac{kQ_1}{r^2} = \frac{9 \times 10^9 \times 4 \times 10^{-6}}{(0.10)^2} = \frac{3.6 \times 10^4}{0.01} = 3.6 \times 10^6 \, \text{N/C} \)
Direction: away from Q₁, i.e., towards Q₂ (let's call this the positive x-direction).
Step 2: Find E₂ at midpoint due to Q₂:
\( E_2 = \frac{k|Q_2|}{r^2} = \frac{9 \times 10^9 \times 4 \times 10^{-6}}{(0.10)^2} = 3.6 \times 10^6 \, \text{N/C} \)
Direction: towards Q₂ (negative charge attracts field lines), which is also in the positive x-direction.
Step 3: Both fields point in the same direction (from Q₁ towards Q₂). Add them:
\( E_{net} = E_1 + E_2 = 3.6 \times 10^6 + 3.6 \times 10^6 = 7.2 \times 10^6 \, \text{N/C} \)
Answer
The net electric field at the midpoint = 7.2 × 10⁶ N/C, directed from the positive charge towards the negative charge.
Example 3 (JEE/NEET Level — Concept Application)
Problem: A uniformly charged thin spherical shell of radius R = 0.5 m carries a total charge Q = 2 μC. Find: (a) the electric field at a point 1 m from the centre (outside), and (b) the electric field at a point 0.2 m from the centre (inside the shell).
Given: R = 0.5 m, Q = 2 μC = 2 × 10⁻⁶ C, r₁ = 1 m (outside), r₂ = 0.2 m (inside).
Step 1 (Outside): By Gauss's Law, a uniformly charged spherical shell behaves like a point charge for points outside it. Apply the Electric Field Formula:
\( E_{outside} = \frac{kQ}{r_1^2} = \frac{9 \times 10^9 \times 2 \times 10^{-6}}{(1)^2} = 18000 \, \text{N/C} = 1.8 \times 10^4 \, \text{N/C} \)
Step 2 (Inside): For a point inside a uniformly charged spherical shell, the enclosed charge is zero. By Gauss's Law:
\( \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} = \frac{0}{\epsilon_0} = 0 \)
Therefore, \( E_{inside} = 0 \, \text{N/C} \).
Step 3: This is a key JEE/NEET result. The electric field inside a conducting or uniformly charged spherical shell is always zero, regardless of the shell's charge.
Answer
(a) Electric field outside at r = 1 m: E = 1.8 × 10⁴ N/C (radially outward). (b) Electric field inside the shell at r = 0.2 m: E = 0 N/C.
CBSE Exam Tips 2025-26
- Memorise the formula in both forms: Write \( E = kQ/r^2 \) and \( E = Q/(4\pi\epsilon_0 r^2) \) side by side. CBSE board papers accept both. We recommend practising unit conversion between N/C and V/m before the exam.
- Direction matters in vector questions: CBSE Class 12 papers frequently ask for the direction of E. Always state whether the field points away from (positive charge) or towards (negative charge) the source.
- Superposition principle is high-yield: Questions involving two or more charges at a point require vector addition of individual fields. Draw a diagram first to identify directions before calculating.
- Gauss's Law shortcuts: For symmetric charge distributions (sphere, cylinder, infinite plane), use Gauss's Law instead of direct integration. This saves time in 3-mark and 5-mark questions.
- Standard values to memorise: k = 9 × 10⁹ N m² C⁻², ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻², and e = 1.6 × 10⁻¹⁹ C. These appear in almost every electrostatics problem in the 2025-26 board exam.
- Units check: Always verify your answer has units of N/C or V/m. A missing or wrong unit costs half a mark in CBSE board exams.
Common Mistakes to Avoid
- Forgetting to square the distance: The most common error is using r instead of r² in the denominator. The Electric Field Formula has r², not r. Double-check before substituting values.
- Confusing E with F: Electric field E = F/q is force per unit charge. Many students substitute the total force F directly instead of dividing by the test charge. Always apply the definition correctly.
- Ignoring vector nature: When two or more charges create fields at the same point, students often add magnitudes directly. You must add fields as vectors, considering both magnitude and direction.
- Wrong unit conversion: Converting μC to C is a frequent error. Remember: 1 μC = 10⁻⁶ C, 1 nC = 10⁻⁹ C, and 1 pC = 10⁻¹² C. Substituting μC directly without conversion gives an answer off by a factor of 10⁶.
- Assuming non-zero field inside a conductor: The electric field inside a conductor in electrostatic equilibrium is always zero. Do not apply the point charge formula for interior points of conductors or charged shells.
JEE/NEET Application of Electric Field Formula
In our experience, JEE aspirants encounter the Electric Field Formula in at least 2–3 questions per paper, spanning electrostatics, capacitors, and motion of charged particles. NEET papers consistently feature 1–2 direct or applied questions on this topic every year.
Pattern 1: Motion of a Charged Particle in a Uniform Electric Field
JEE Main frequently asks about a charged particle (mass m, charge q) entering a uniform electric field E between parallel plates. The acceleration is \( a = qE/m \). The motion is analogous to projectile motion. Students must combine kinematics equations with the Electric Field Formula to find deflection, time of flight, and exit velocity. This is a multi-concept question worth 4 marks in JEE Main.
Pattern 2: Electric Field on the Axis of a Dipole
JEE Advanced and NEET both test the dipole field formula. At a point on the axial line at distance r from the centre, \( E_{axial} = 2kp/r^3 \). At a point on the equatorial line, \( E_{eq} = kp/r^3 \). The ratio E_axial : E_eq = 2 : 1 is a frequently tested result. Our experts suggest memorising this ratio directly, as it appears in MCQ options regularly.
Pattern 3: Application of Gauss's Law
Both JEE Advanced and NEET test the electric field due to symmetric charge distributions using Gauss's Law. Key results to memorise:
- Infinite line charge: \( E = \frac{\lambda}{2\pi\epsilon_0 r} \)
- Infinite plane sheet: \( E = \frac{\sigma}{2\epsilon_0} \)
- Uniformly charged sphere (outside, r > R): \( E = \frac{Q}{4\pi\epsilon_0 r^2} \)
- Uniformly charged sphere (inside, r < R): \( E = \frac{Qr}{4\pi\epsilon_0 R^3} \)
In our experience, JEE aspirants who practise these four results as standard templates solve Gauss's Law questions in under 90 seconds, saving crucial time for harder problems.
FAQs on Electric Field Formula
Explore more related topics on ncertbooks.net to strengthen your electrostatics preparation. Read our detailed article on Capacitance Formula to understand how electric fields relate to charge storage in capacitors. For a broader understanding of charge interactions, visit our Physics Formulas hub, which covers all NCERT Class 12 Physics topics. You can also revise the Refractive Index Formula for your optics preparation. For official CBSE syllabus details, refer to the CBSE official website.